The availability of S Chand ISC Maths Class 12 Solutions Chapter 8 Differentiation Ex 8(i) encourages students to tackle difficult exercises.
S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(i)
Find \(\frac { dy }{ dx }\) :
Question 1.
If x = ct, y = \(\frac { c }{ t }\).
Solution:
Given x = ct … (1)
and y = \(\frac { c }{ t }\) … (2)
DifF. eqn. (1) & eqn. (2) w.r.t. t ; we have
\(\frac { dx }{ dt }\) = c & \(\frac { dy }{ dt }\) = – \(\frac { c }{ t² }\)
∴ \(\frac{d y}{d t}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=-\frac{\frac{c}{t^2}}{c}=-\frac{1}{t^2}\)
Question 2.
x = \(\frac{1}{1-t^2}\), y = 1 + t².
Solution:
Question 3.
x = a\(\left(\frac{1+t^2}{1-t^2}\right)\), y = b\(\left(\frac{2 t}{1-t^2}\right)\)
Solution:
Question 4.
If x = a (θ + sin θ), y – a{ 1 – cos θ)
Solution:
Given x – a (0 + sin θ) …(1)
& y = a{ 1 – cos θ) …(2)
Diff. eqn. (1) & eqn. (2) w.r.t. θ ; we have
Question 5.
If x = b sin² θ and y = a cos²θ.
Solution:
Given x = b sin² θ …(1)
& y = a cos²θ …(2)
Diff. eqn. (1) & eqn. (2) w.r.t. t ; we have
Question 6.
If x = a cos³ t, y = a sin³ t.
Solution:
Given x = a cos³ t …(1)
& y = a sin³t …(2)
Diff. eqn. (1) & eqn. (2) w.r.t. t ; we have
Question 7.
x = a(cos t + t sin t), y = a(sin t – t cos t).
Solution:
Given x = a(cos t + t sin t) …(1)
& y = a(sin t – t cos t) …(2)
Diff. eqn. (1) & eqn. (2) w.r.t. t ; we have
Question 8.
x = 3 cos t – 2 cos3 t, y = 3 sin t – 2 sin³t.
Solution:
Given x = (3 cos t – 2 cos³ t) …(1)
& y = 3 sin t – 2 sin³t …(2)
Diff. eqn. (1) & eqn. (2) w.r.t. t ; we have
Question 9.
Find \(\frac { dy }{ dx }\) when x = log t, y = sin t.
Solution:
Given x = log t …(1)
& y = sin t …(2)
Diff. eqn. (1) & eqn. (2) w.r.t. t ; we
\(\frac{d x}{d t}=\frac{1}{t} \& \frac{d y}{d t}=\cos t\)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\cos t}{\frac{1}{t}}=t \cos t\)
Question 10.
If x = a (cos θ + log tan \(\frac { θ }{ 2 }\)) and y = a sin θ, find \(\frac { dy }{ dx }\) at θ = \(\frac { π }{ 3 }\) and θ = \(\frac { π }{ 4 }\).
Solution:
Given x = a (cos θ + log tan \(\frac { θ }{ 2 }\) ) … (1)
& y = a sin θ … (2)
Diff. eqn. (1) & eqn. (2) w.r.t. θ ; we have
